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Musya8 [376]
4 years ago
6

What is the modulus of resilience for a tensile test specimen with a nearly linear elastic region if the yield strength is 500MP

a at a strain of 0.003
Engineering
2 answers:
Temka [501]4 years ago
4 0

Answer:

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

Explanation:

The Modulus of resilience can be derived from the area under the elastic portion of the stress-strain diagram. It can be written mathematically as;

U = 1/2 × yield strength × yield strain.

Where U is the modulus of resilience.

Given;

Yield strength = 500MPa

Yield strain = 0.003

Substituting the given values we have;

U = 1/2 × 500MPa × 0.003

U = 0.75 MPa

Therefore, the modulus of resilience for the tensile test specimen is 0.75MPa

ella [17]4 years ago
3 0

Answer:

The modulus of resilience is 166.67 MPa

Explanation:

Modulus of resilience is given by yield strength ÷ strain

Yield strength = 500 MPa

Strain = 0.003

Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa

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Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb
Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = \sqrt{400^2 + 650^2 - 2(400)(650)cos30}

      = \sqrt{160000 + 422500 - 80210}

      = 708.72 Ib

using the sine law to determine the design angle

\frac{sin}{400}  = \frac{sin 30}{Fac}

hence ∅ = sin^{-1} (\frac{sin 30 *400}{708.72} )

              = sin^{-1} 0.0871 =  4.9968 ≈ 5⁰

4 0
3 years ago
6. Question
valkas [14]

Answer:

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Amiraneli [1.4K]
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3 years ago
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Answer:

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Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.
Sedbober [7]

Answer:

True

Explanation:

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